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Q.

A missile of mass M moving with velocity v in free space explodes into two parts. After the explosion one of the parts of mass m falls vertically down. What will be the velocity of the other part?

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a

Mv / m

b

Mv / (M – m)

c

(M – m) v / M

d

mv / M - m

answer is B.

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Detailed Solution

The velocity of the other part is

\frac{Mv}{M – m}

.
According to the law of conservation of linear momentum, a system's total momentum remains constant. If the missile detonates into a component with mass m and velocity v, the mass of the remaining part will be
Mass of the second part =

M - m .

If its speed is v, then the conservation principle states that the total momentum before explosion will be equal to the sum of the momentums of the individual parts.

MV = (M - m) v + m \times 0

= > v = M v / (M - m)

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